Problem: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $n \neq 0$. $p = \dfrac{8(3n + 5)}{4} \times \dfrac{4n}{21n^2 + 35n} $
When multiplying fractions, we multiply the numerators and the denominators. $p = \dfrac{ 8(3n + 5) \times 4n } { 4 \times (21n^2 + 35n) } $ $ p = \dfrac {4n \times 8(3n + 5)} {4 \times 7n(3n + 5)} $ $ p = \dfrac{32n(3n + 5)}{28n(3n + 5)} $ We can cancel the $3n + 5$ so long as $3n + 5 \neq 0$ Therefore $n \neq -\dfrac{5}{3}$ $p = \dfrac{32n \cancel{(3n + 5})}{28n \cancel{(3n + 5)}} = \dfrac{32n}{28n} = \dfrac{8}{7} $